## College Algebra (10th Edition)

$A^{-1}=\left[\begin{array}{rrr} {\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\ {\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\ {-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$
To find the inverse of an $n$ by $n$ nonsingular matrix $A,$ proceed as follows: STEP 1: Form the matrix $\left[A|I_{n}\right]$ STEP 2: Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form. STEP 3: The reduced row echelon form of $\left[A|I_{n}\right]$ will contain the identity matrix $I_{n}$ on the left of the vertical bar; the $n$ by $n$ matrix on the right of the vertical bar is the inverse of $A.$ If the identity matrix can not be obtained on the left, A has no inverse. ---- STEP 1 $\left[A|I_{n}\right] = \left[\begin{array}{rrr|rrr} {3}&{3}&{1}&{1}&{0}&{0}\\ {1}&{2}&{1}&{0}&{1}&{0}\\ {2}&{-1}&{1}&{0}&{0}&{1}\end{array}\right]\qquad \left(\begin{array}{l} r_{1}\leftrightarrow r_{2}.\\ .\\ . \end{array}\right)$ STEP 2 $\rightarrow\left[\begin{array}{rrr|rrr} {1}&{2}&{1}&{0}&{1}&{0}\\ {3}&{3}&{1}&{1}&{0}&{0}\\ {2}&{-1}&{1}&{0}&{0}&{1}\end{array}\right] \qquad \left(\begin{array}{l} .\\ R_{2}=r_{2}-3r_{1}.\\ R_{3}=r_{3}-2r_{1}. \end{array}\right)$ $\rightarrow\left[\begin{array}{rrr|rrr} {1} & {2} & {1} & {0} & {1} & {0} \\ {0} &{-3}&{-2}&{1} &{-3} &{0}\\ {0} &{-5}&{-1}&{0} &{-2} &{1}\end{array}\right] \qquad \left(\begin{array}{l} .\\ R_{2}=-2r_{2}+r_{3}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{rrr|rrr} {1} & {2} & {1} & {0} & {1} & {0} \\ {0} &{1} &{3} &{-2} &{4} &{1}\\ {0} &{-5}&{-1}&{0} &{-2} &{1}\end{array}\right] \qquad\left(\begin{array}{l} R_{1}=r_{1}-2r_{2}.\\ .\\ R_{3}=r_{3}+5r_{2}. \end{array}\right)$ $\rightarrow\left[\begin{array}{rrr|rrr} {1} & {0} & {-5} & {4} & {-7} & {-2} \\ {0} &{1} &{3} &{-2} &{4} &{1}\\ {0} &{0} &{14} &{-10} &{18} &{6}\end{array}\right] \qquad \left(\begin{array}{l} .\\ .\\ \div 14. \end{array}\right)$ $\rightarrow\left[\begin{array}{rrr|rrr} {1} & {0} & {-5} & {4} & {-7} & {-2} \\ {0} &{1} &{3} &{-2} &{4} &{1}\\ {0} &{0} &{1} &{-5/7} &{9/7} &{3/7}\end{array}\right] \qquad \left(\begin{array}{l} R_{1}=r_{1}+5r_{3}.\\ R_{2}=r_{2}-3r_{3}.\\ . \end{array}\right)$ $\rightarrow\left[\begin{array}{ccc|rrr} {1}&{0}&{0}&{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\ {0}&{1}&{0}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\ {0}&{0}&{1}&{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$ STEP 3 $A^{-1}=\left[\begin{array}{rrr} {\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\ {\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\ {-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$